Compact Integral operator? I have stumbled upon an interesting linear operator. I am not sure if it is compact:
$$
T: L^2([0, 1]) \rightarrow L^2([0, 1]), \quad (Tf)(x) := \int^1_0 (x-y)^2 f(y)~\mathrm{d}y.
$$
I tried to use the following argument: Let $(f_k)_{k \in \mathbb{N}} \subseteq L^2([0, 1])$ be bounded and let $(f_{n_k})_{k \in \mathbb{N}} \subseteq L^2([0, 1])$ be a subsequence with weak limit $f \in L^2([0, 1])$. Then:
$$
\lVert Tf_{n_k} - Tf \rVert_{L^2([0, 1])}^2 = \int_0^1 \left \lvert \int^1_0 (x-y)^2 (f_{n_k}(y) - f(y))~\mathrm{d}y \right \rvert^2 ~\mathrm{d}x \leq \sup_{x \in [0, 1]} \left \lvert \int^1_0 (x-y)^2 (f_{n_k}(y) - f(y))~\mathrm{d}y \right \rvert^2
$$
I can still prove that this $\sup$ is attained on some $x_{n_k} \in [0, 1]$ but this does not help me because of the dependancy on $n_k$. Otherwise I would have used weak convergence at that point.
Is there an easier way or this is operator even compact?
 A: It's a finite rank operator and thus compact: $$(Tf)(x)=A\,x^2-B\,x+C,$$ where
$A=\int^1_0f(y)\,dy, B=2\int^1_0y\,f(y)\,dy, C=\int^1_0y^2\,f(y)\,dy$, so the image of a bounded set is a bounded subset of a 3-dimensional subspace of $L^2([0, 1])$.
A: If $k(x,y)=a(x)b(y)$, where $a$ and $b$ are continuous functions on $[0,1]$, then
$$
\int_0^1 k(x,y) f(x)\, dx = \left(\int_0^1 a(x) f(x)\, dx\right) b(y) = \langle a,f\rangle b(y).
$$
Therefore the integral operator $T_k$, given by the integral kernel $k$,  satisfies $T_k(f) =  \langle a,f\rangle b$,
and, as a consequence,
$T_k$ is seen to be a rank one operator.  If instead,
$$
  k(x,y)=\sum_{i=1}^na_i(x)b_i(y),
  \tag{1}
  $$
where $a_i$ and $b_i$ are again continuous functions on $[0,1]$, the corresponding integral operator
satisfies
$$
  T_k(f) =  \sum_{i=1}^n \langle a_i,f\rangle b_i, \quad \forall f\in  L^2[0,1],
  $$
so the rank of $T_k$ is at most $n$,  hence finite.
If we next assume that $k$ is a uniform limit of functions $k_n$ of the form (1), it is not hard to see that
$T_{k_n}\to T_k$, in operator norm, so $T_k$ is a compact operator since it is  the limit of finite rank operators.
The Stone-Weierstrass Theorem may be easily invoked to show that the subset  of $C([0,1]\times[0,1])$ formed by all
functions of the form (1) is dense, so the conclusion of the paragraph above holds for any  continuous $k$.   This proves:
Theorem.  Any integral operator on $L^2[0,1]$ with a continuous integral kernel is compact.
